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3 years ago

8 people eat at Charlie Brown's for Thanksgiving. How many different seating

Mathematics

2 answers:

never [62]3 years ago

4 0

Answer: 40,320

Step-by-step explanation: Let's say that there is person A,B,C,D,E,F,G,H. and 8 chairs. For the first chair, 8 different people could potentially sit in it, making 8 different possibilities. No matter who sits there, the logic follows the next table. However, since one person is sitting in the first chair, there are 7 different possibilities about who would be sitting in the second chair. If you multiply the two together, there are 56 different possibilities just for the first and second chair. For the third chair, there are 6 different possibilities about whom could sit. Multiply 56*6 and you get 336 possibilities. Keep multiplying out and you get a grand total of 40,320 different arrangements!

Alinara [238K]3 years ago

4 0

Answer:

5040

Step-by-step explanation:

The answer is (8 - 1)!

The reason is that the first person is fixed. Any place is the same as any other place. After that there are 7! choices for the next position.

Answer: 5040

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3 years ago

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Margaret [11]

Answer:

The explicit form is $a_{n}=162(2/3)^{n-1}$

Step-by-step explanation:

The explicit form of a geometric sequence is given by:

$a_{n}=ar^{n-1}$

where an is the nth term, a is the first term of the sequence and r is the common ratio.

In this case:

a=162

The value of the common ratio is obtained by dividing one term by the previous term.

For the first and second terms:

108/162=2/3

For the second and third terms (In order to prove that 2/3 is the common ratio)

72/108=2/3

Therefore:

r=2/3

Replacing a and r in the formula:

$a_{n}=162(2/3)^{n-1}$

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